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Rasch models
Karl Bang Christensen
Dept. of BiostatisticsUniv. of Copenhagen
http://publicifsv.sund.ku.dk/~kach/scaleval_IRT
Karl Bang Christensen Rasch models
IRT model for dichotomous item
specifies that the probability of response pattern xv = (xvi )i=1,...,I
is
P(xv ) =
∫ I∏i=1
(Pi (θ)xvi (1− Pi (θ))(1−xvi )
)φ(θ)dθ
where is Pi (θ) is the probability of a correct response on item i asa function of the trait θ, and φ is the population distribution for θ.
Last week
Pi (θ) 2PL, φ(θ) standard normal distribution (mean=zero, SD=1).
This week
Pi (θ) 1PL, no assumption distributional assumptions are needed.
Karl Bang Christensen Rasch models
Rasch model: Probabilities for item i
person location θ, location βi and scale αi of item i .
P(Xvi = 1|θv = θ) =exp(θ − βi )
1 + exp(θ − βi )
The ’trick’
P(Xvi = xi ,Xvi ′ = xi ′ |Rv = xi + xi ′ , θ)
where Rv = Xvi + Xvi ′ is independent of θ
Karl Bang Christensen Rasch models
Model framework, Assumptions
Unidimensional latent variable θv responsible for all correlationbetween the observed items items X = (Xi )i∈I , covariatesY = (Y1,Y2, ..)
(i) Θ unidimensional
(ii) Monotonous relationship between Θ and Xi
(iii) No differential item functioning (DIF) Xi ⊥ Yj |Θ(iv) Local independence: Xi ⊥ Xj |ΘRasch model (=1PL) further requirements
sufficiency, specific objectivity, exchangeability (invariance)
Karl Bang Christensen Rasch models
Invariance
P(X1 = 1|θ) P(X1 = 0|θ)
P(X2 = 1|θ) P(X2 = 0|θ)
=
exp(θ−β1)
1+exp(θ−β1)1
1+exp(θ−β1)
exp(θ−β2)1+exp(θ−β2)
11+exp(θ−β2)
Odds
P(X1=1|θ)P(X1=0|θ)
P(X2=1|θ)P(X2=0|θ)
=
exp(θ − β1)
exp(θ − β2)
Comparison OR = exp(β2 − β1) independent of θ
Karl Bang Christensen Rasch models
Sufficiency
Local independence gives us
P(X = x|θ) =I∏
i=1
exp(xi (θ − βi ))1 + exp(θ − βi )
=exp(
∑Ii=1 xi (θ − βi ))∏I
i=1(1 + exp(θ − βi ))
=exp(tθ −
∑Ii=1 xiβi )∏k
i=1(1 + exp(θ − βi ))=
exp(tθ −∑I
i=1 xiβi )
K(θ,β)
where t =∑k
i=1 xi is the sum score.
Karl Bang Christensen Rasch models
Sufficiency
The sum score t =∑
i xi is sufficient for θ.
This means that
P(X1 = h1, . . . ,XI = hI |X1 + · · ·+ XI = t)
does not depend on θ.
We can use this to estimate item parameters without makingassumptions about the distribution of the latent variable.
Karl Bang Christensen Rasch models
Estimation
Marginal likelihood
normal distribution (mean zero, SD σ)
LM(η) =N∏
v=1
∫P(Xv = xv |θv = θ)ϕσ(θ) (1)
consistent estimates if normal distribution is correct.
Conditional likelihood
no distributional assumptions
LC (η) =N∏
v=1
P(Xv = xv |Rv = rv ) (2)
conditionally consistent estimates1
1Andersen. Journal of the Royal Statistical Society B, 1970, 32:283-301.Karl Bang Christensen Rasch models
Estimation in SAS
Marginal likelihood
PROC IRT
macro %rasch mml.sas2
Graphics: item characteristic curves (ICC’s), person-item locationmaps, item and test information functions. Goodness-of-fit plots.
2Christensen, Olsbjerg. Marginal maximum likelihood estimation inpolytomous Rasch models using SAS. Pub. Inst. Stat. Univ. Paris, vol. 57,fasc. 1-2, 69-84, 2013.
Karl Bang Christensen Rasch models
Estimation in SAS
Conditional likelihood
macro %rasch cml.sas3
Graphics: item characteristic curves (ICC’s), person-item locationmaps, item and test information functions. Goodness-of-fit plots.
3Christensen. Conditional maximum likelihood estimation in polytomousRasch models using SAS, ISRN Computational Mathematics, vol. 2013, ArticleID 617475, 8 pages, 2013 http://doi.org/10.1155/2013/617475
Karl Bang Christensen Rasch models
ADL six months after breast cancer surgery
17 items (ADL=Activities of Daily Living)
dichotomized: (0=no problems, 1=problems)Karl Bang Christensen Rasch models
ADL six months after breast cancer surgery
%l e t i t ems=Q6m 1 Q6m 2 Q6m 3 Q6m 4 Q6m 5 Q6m 6 Q6m 7 Q6m 8 Q6m 9 Q6m 10 Q6m 11 Q6m 12 Q6m 13 Q6m 14 Q6m 15 Q6m 16 Q6m 17 ;∗ r ead ADL data ;data s a s u s e r .ADL;
f i l e n ame dat u r l ’ h t tp : // b i o s t a t . ku . dk/˜kach/ s c a l e v a l IRT/ADL. t x t ’ ;i n f i l e dat f i r s t o b s =2;i n pu t i d &i t ems ;
run ;
Karl Bang Christensen Rasch models
Dichotomous Rasch model for ADL items
Item i taking values 0,1:
P(Xvi = x |θv = θ) =exp(x(θ − βi ))
1 + exp(θ − βi )
proc i r t data=s a s u s e r .ADL p l o t s =( i c c i i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=ra s ch ;
run ;
PROC IRT uses marginal likelihood (assumes θ ∼ N(0, σ2))
parameters
β1, . . . , β17, σ
Karl Bang Christensen Rasch models
Dichotomous Rasch model for ADL items
Item i taking values 0,1:
P(Xvi = x |θv = θ) =exp(xα(θ − βi ))
1 + exp(α(θ − βi ))
proc i r t data=s a s u s e r .ADL p l o t s =( i c c t i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=onep ;
run ;
PROC IRT uses marginal likelihood (assumes θ ∼ N(0, 1))
parameters
β1, . . . , β17, α
Karl Bang Christensen Rasch models
Dichotomous Rasch model for ADL items
Item Parameter Estimate s.e.
Q6m 1 Difficulty 3.36981 0.21681Slope 1.00000
Q6m 2 Difficulty 1.54327 0.15923Slope 1.00000
Karl Bang Christensen Rasch models
Dichotomous Rasch model for ADL items
interpretation of βi is the usual (P = 12)
Karl Bang Christensen Rasch models
Dichotomous Rasch model for ADL items
Item i taking values 0,1:
P(Xvi = x |θv = θ, βi1) =exp(x(θ − β1))
1 + exp(θ − β1)
proc i r t data=s a s u s e r .ADL p l o t s =( i c c i i c ) ;va r Q6m 1−Q6m 17 ;model Q6m 1−Q6m 17 / r e s f u n c=ra s ch ;
run ;
Karl Bang Christensen Rasch models
Rasch model in SAS, CML
Include macros
%l e t u r l=h t t p s : // raw . g i t h u bu s e r c o n t e n t . com/Ka r lBangCh r i s t en s en /Rasch/master ;f i l e n ame r u r l ”&u r l / r a s ch i n c l u d e a l l . s a s ” ;%i n c l u d e r ;
Item information
data i n ;i n pu t i tem no i tem name $ i tem t e x t $ max group ;d a t a l i n e s ;1 Q6m 1 x 1 12 Q6m 2 x 1 23 Q6m 3 x 1 34 Q6m 4 x 1 45 Q6m 5 x 1 56 Q6m 6 x 1 67 Q6m 7 x 1 78 Q6m 8 x 1 89 Q6m 9 x 1 910 Q6m 10 x 1 1011 Q6m 11 x 1 1112 Q6m 12 x 1 1213 Q6m 13 x 1 1314 Q6m 14 x 1 1415 Q6m 15 x 1 1516 Q6m 16 x 1 1617 Q6m 17 x 1 17;run ;
Karl Bang Christensen Rasch models
Rasch model in SAS, CML
call macro
%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;
%ra s ch CML(data=s a s u s e r .ADL,i tem names=in ,out=CML) ;
takes a long time for big tables (contingency table has 217 cells).
Karl Bang Christensen Rasch models
Rasch model in SAS, MML
call macro
%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;
%ra s ch MML(data=s a s u s e r .ADL,i tem names=in ,out=MML) ;
takes a long time for big data sets.
Karl Bang Christensen Rasch models
Does the Rasch model fit data? Overall test of the model4
Divide sample into G groups: Lg likelihood in group g
CLR = −2 log
(L∏G
g=1 Lg
)
asymptotically χ2 on (G − 1) · (I − 1) degrees of freedom (fordichotomous items). Note that this uses the fact that we do nothave to assume anything about the distribution
4Andersen. Psychometrika, vol. 38, 123-140, 1973.Karl Bang Christensen Rasch models
Does the Rasch model fit data? Overall test of the model
Two data sets G = 1 (score 0-2), G = 2 (score 3-17)
data ADL;s e t s a s u s e r .ADL;s c o r e=sum( o f &i t ems ) ;nm=nmiss ( o f &i t ems ) ;
run ;data ADL;
s e t ADL;i f nm=0;
run ;data ADL1 ;
s e t ADL;i f s c o r e i n ( 0 , 1 , 2 , 3 ) ;
run ;data ADL2 ;
s e t ADL;i f s c o r e > 3 ;
run ;
Karl Bang Christensen Rasch models
Does the Rasch model fit data? Overall test of the model
Fit Rasch model in each data set
%ra s ch data (data=ADL,
i tem names=i n ) ;%ra s ch cml (
data=ADL,i tem names=in ,out=cml ) ;
%ra s ch data (data=ADL1 ,
i tem names=i n ) ;%ra s ch cml (
data=ADL1 ,i tem names=in ,out=cml1 ) ;
%ra s ch data (data=ADL2 ,
i tem names=i n ) ;%ra s ch cml (
data=ADL2 ,i tem names=in ,out=cml2 ) ;
Karl Bang Christensen Rasch models
Does the Rasch model fit data? Overall test of the model
Compare fit of original Rasch model L with combined fit in eachdata set L1 · L2proc s q l ;
s e l e c t v a l u e i n t o : l 1 from cml1 l o g l ;s e l e c t v a l u e i n t o : l 2 from cml2 l o g l ;s e l e c t v a l u e i n t o : l from cml l o g l ;
q u i t ;data l r t ;
l r t =(& l 1+& l2−& l ) ;d f =16;p=1−cd f ( ’ c h i s q u a r e d ’ , l r t , d f ) ;
run ;p roc p r i n t data= l r t round noobs ;run ;
by compare log likelihood values
χ2 = 48.2(df = 16)
model is rejected
Karl Bang Christensen Rasch models
Multiple groups analysis 5
G = 1 X1 X2
X3
:
Θ
KK EE
::
--
&&
:
X16
X17
G = 2 X1 X2
X3
:
Θ
KK EE
::
--
&&
:
X16
X17
5Andersen. Psychometrika, vol. 38, 123-140, 1973.Karl Bang Christensen Rasch models
Does the Rasch model fit data? Individual item fit
ResidualsRvi = Xvi − Evi
where Xvi is the response of person v to item i and Evi expectedresponse of person v to item i .
Two issues to discuss
(i) How to calculate the expected values
(ii) How to summarize residuals
Karl Bang Christensen Rasch models
The conditional item characteristic curve
The item characteristic curve is the item response function
θ 7→ P(Xvi = x |θv = θ).
In the Rasch model: simple expression for a conditional version ofthis
r 7→ P(Xvi = x |Rv = r) =exp(−βi )γr−1(β1, . . . , βi−1, βi+1, . . . , βI )
β
Karl Bang Christensen Rasch models
(i) How to calculate the expected values
Expected response Evi :ideal situation: Known person locations
Evi = P(Xvi = 1|Θv = θv ) =exp(θv − βi )
1 + exp(θv − βi )
less than ideal situation: Person locations are estimated
Evi = P(Xvi = 1|Θv = θ̂v ) =exp(θ̂v − β̂i )
1 + exp(θ̂v − β̂i )
Two problems
Person locations are estimated with error
Formula is incorrect
Commercial Rasch model software ignores these problems ..
Karl Bang Christensen Rasch models
Standard evaluation of individual item fit
using
Ziv =Xiv − E (Xiv |Θv = θ̂v )√
V (Xiv |Θv = θ̂v )
Fit statistics like
OUTFITi =1
N
N∑v=1
Z 2iv
with no established null distribution are used. Early Raschliterature claims of χ2 distribution.6
6Wright, Stone (1979). Mesa Press, Chicago, USA.Karl Bang Christensen Rasch models
Wilson-Hilferty cube-root transformation
It has been suggested that the Wilson-Hilferty (1931) cube-roottransformation
ti = (OUTFIT1/3i − 1)
3
V (OUTFITi )+
V (OUTFITi )
3
has an approximate t distribution and that
FitResidi =f (log(N · OUTFITi )− log(f ))√
V (N · OUTFITi )
with f = (Nk − N − k + 1)/k, has a symmetrical distribution withmean zero and variance one.
Karl Bang Christensen Rasch models
Commercial software
WINSTEPS
OUTFIT, t-transformed OUTFIT
INFIT (weighted version of OUTFIT), t-transformed INFIT
RUMM
item χ2 fit statistic
item ANOVA fit statistic
item FitResidi
Karl Bang Christensen Rasch models
RUMM
groups respondents in G ’class intervals’ based on the estimatedperson locations θ̂v .
still relies on θ̂1, θ̂2, . . .
Item χ2 fit statistic
χ2(Xi ) =∑g
(∑v∈Vg
Xvi −∑
v∈VgE (Xvi )
)2∑v∈Vg
V (Xvi )
where Vg denotes the set of respondents in class interval g . ItemANOVA fit statistic uses ANOVA based on the grouping
Ziv = µg(v) + ERROR
Karl Bang Christensen Rasch models
Item fit statistics in SAS
Fit model
%ra s ch data (data=s a s u s e r .ADL,i tem names=i n ) ;
%ra s ch CML(data=s a s u s e r .ADL,i tem names=in ,out=CML) ;
creates output data sets with information.Estimate person locations
%ra s ch ppar (DATA=s a s u s e r .ADL,ITEM NAMES=in ,DATA IPAR=CML ipa r ,out=pp cml ) ;
also creates output data sets with information
Karl Bang Christensen Rasch models
Item fit statistics in SAS
Compute item fit statistics
%ra s ch i t e m f i t (DATA=s a s u s e r .ADL,ITEM NAMES=in ,DATA IPAR=cml i p a r ,DATA POPPAR=pp cml outdata ,NCLASS=3,OUT=f i t cm l ) ;
(takes a long time). We specify that we want three class intervals.Output data sets ’fitcml chisq’, ...
Karl Bang Christensen Rasch models
Exercise 10: Item fit statistics in SAS
Use the 11 symptom items from the colitis data set.
1 Evaluate item fit using PROC IRT
2 Evaluate item fit using SAS macros
use
http://publicifsv.sund.ku.dk/~kach/scaleval_IRT/exerc10.sas
Karl Bang Christensen Rasch models
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